The interionic dist in FCC=edge/square root of 2.But in NaCl (FCC) dist between Na+ & Cl- is a/2
Asked by | 9th Apr, 2008, 10:58: AM
Anions ccp (fcc). Radius Na+ = 1.02Å, radius Cl- = 1.81Å; radius ratio = 0.563.
Therefore Na octahedral.
1 octahedral / anion therefore 100% octahedral sites are filled.
Coordination # Na = 6; coordination # Cl = 6.
For an fcc lattice there are 4 lattice points per cell, the motif in this case is a Na cation and a Cl anion. Therefore the cell contents are 4 Na cations + 4 Cl anions.
Lattice parameter and Density calculations.
The cation and its nearest neighbor anions are, by definition in contact.
Therefore the cell edge = 2r(Na+) + 2r(Cl-) = 2(1.02) + 2(1.81) = 5.66Å
The Coulombic interactions involve all the surrounding ions:
6 Cl- at a distance d
12 Na+ at a distance d√2
8 Cl- at a distance d√3
6 Na+ at a distance d√4 = 2d
24 Cl- at a distance d√5
and so on…
Thus we have
UC = Z+Z-e2 ( 6 – 12/√2 + 8/√3 – 6/√4 + 24/√5 - ...)
The infinite series can be summed, and Madelung’s constant for the NaCl structure type MNaCl = 1.748.
Answered by | 20th May, 2008, 09:09: AM
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